Solution:

The divisibility condition is equivalent to $b-a$ being divisible by both $a$ and $a+1$, or, equivalently (since these are relatively prime), by $a(a+1)$. Any $b$ satisfying the condition is automatically $\geq a$, so it suffices to count the number of values $b-a \in \{1-a, 2-a, \ldots, 60-a\}$ that are divisible by $a(a+1)$ and sum over all $a$. The number of such values will be precisely $60 / [a(a+1)]$ whenever this quantity is an integer, which fortunately happens for every $a \leq 5$; we count:

$a=1$ gives 30 values of $b$;
$a=2$ gives 10 values of $b$;
$a=3$ gives 5 values of $b$;
$a=4$ gives 3 values of $b$;
$a=5$ gives 2 values of $b$;
$a=6$ gives 2 values ($b=6$ or $48$);
any $a \geq 7$ gives only one value, namely $b=a$, since $b>a$ implies $b \geq a+a(a+1)>60$.

Adding these up, we get a total of 106 pairs.